2j2
CHAPTER
8 Chemical
Bonding
I:
Basic
Concepts
r
Think
About
It Counting the total
number
of
valence electrons should
be relatively simple to do, but it
is
often done hastily and is therefore
a potential source
of
error in this
type
of
problem. Remember that
the number
of
valence electrons for
each element is equal to the group
number
of
that element.
,

M
ultimedia
Chemical
Bonding
- formal ch
arge
ca
lculation
s.
Draw the Lewis structure for carbon disulfide (CS
2
).
Strategy
Use the procedure described in steps 1 through 6 on page 291 for drawing Lewis structures.
Setup
Step 1:
C and S have identical electronegativities.
We
will draw the skeletal structure with the
unique atom, C, at the center.
S-C-S
Step 2: The total number
of
valence electrons is 16, six from each S atom and four from the C
atom [2(6)
+ 4 = 16].
Step
3:
Subtract four electrons to account for the bonds in the skeletal structure, leaving us 12
electrons to distribute.

Step 4:
St
ep 5:
Step
6:
Distribute the 12 remaining electrons as three lone pairs on each S atom.
• • • •
:S-C-S:
••
••
There are no electrons remaining after step 4, so step 5 does not apply.
To complete carbon's octet, use one lone pair from each S atom to make a double bond to
the C ato
m.
Solution
• •
:S=C=S:
• •
Practice
Problem
A Draw the Lewis structure for
NF
3
.
Practice
Problem
B Draw the Lewis structure for
CI0
3
.

Checkpoint 8.5
Drawing Lewis Structures
8.5.1
Identify the correct Lewis structure for
formic acid
(H
COO
H
).
• • • • • •
a)
H-C-O-O-H
• • • • • •
• • • •
b)
H-O=C=O-H
• • • • • •
c) H
-O-C-O-H
• • • • • •
'0'
• •
II

d)
H-C-O-H


'0'
• •


I
e)
:O-C-H

I
H
8.5.2
Identify the correct Lewis structure for
hydrogen peroxide (H
2
0
2
) .

"
a)
H-O=O-H
b)
H-O=O-H
•• ••
c)
H-O-O-H
• • • •
d)
H=O=O=H
• • • •
e)
H-H-O-O:
••

• •
Lewis Structures and Formal Charge
So far you have learned two different methods
of
electron "bookkeeping." In Chapter 4, you learned
about oxidation numbers
[
~~
Sectio
n 4.4], and in Section 8.4, you learned how to calculate par-
tial charges. There is one additional commonly used method
of
electron bookkeeping namely,
formal charge, which can be used to determine the most plausible Lewis structures when more
than one possibility exists for a compound. Formal charge is determined by comparing the number
of
electrons associated with
an
atom in a Lewis structure with the number
of
electrons that would
be associated with the isolated atom. In an isolated atom, the number
of
electrons associated with
the atom is simply the number
of
valence electrons. (As usual, we need not be concerned with the
core electrons.)
To
determine the number

of
electrons associated with an atom in a Lewis structure, keep in
mind the following:
SECTION 8.6 Lewis Structures and Formal Charge 293
o All the atom's nonbonding electrons are associated with the atom.
o
Half
of
the atom's bonding electrons are associated with the atom.
formal charge
= valence electrons - associated electrons
Equation
8.
2
An
atom's formal charge is calculated as follows: We can illustrate the concept
of
formal charge
using the
ozone
molecule
(0
3
) ,
Use the step-by-step method for drawing Lewis structures to draw
the Lewis structure for ozone, and then determine the formal charge
on
each 0 atom by subtract-
ing the
number

of
associated electrons from the number
of
valence electrons.
2 unshared
+ 6
s
h~red
= 5 e-
4 unshared + 4 s
har
ed = 6
e-
2
Valence
e-
e- associated with atom
Difference (formal charge)
6
un
shared + 2 s
\ared
= 7
e-
6 6 6
6 5 7
o + 1
-1
\ /
:0=0-0:

• • •
As with oxidation numbers, the
sum
of
the formal
char
ges
mu
st equal the overall charge on
the species
[I
~~
Section
4.4].
Because 0
3
is a molecule, its formal charges
mu
st s
um
to zero. For
ions, the formal charges
mu
st s
um
to the overall charge on the ion.
Formal charges do
not
repres
ent

actual charges on atoms in a molecule. In the 0
3
molecule,
for example, there is no evidence that the central atom bears a net
+ 1 charge or that one
of
the
terminal atoms bears a
- 1 charge. Assigning formal charges to the atoms in the Lewis structure
merely helps us keep track
of
the electrons inv
ol
ved in bonding in the molecule.
Sample Problem 8.7 lets you practice determining formal charges.
The widespread use
of
fertilizers has resulted in the contamination of some groundwater with
nitrates, which are potentially harmful. Nitrate
to
xicity is due primarily to its conversion in the body
to nitrite (
N0
2
),
which interferes with the ability
of
hemoglobin to transport oxyge
n.
Determine the

formal charges on each atom in the nitrate ion
(NO
))
.
Strategy
Use steps 1 through 6 on page 291 for drawing Lewis stuctures to draw the L
ew
is structure
of
NO) . For each atom, subtract the associated electrons from the valence electrons.
Setup

'0'
• •

I .
:O-N=O:
••
•
The N atom has five valence
eJ
ectrons and four associated electrons (one from each single bond and
two from the double bond). Each singly bonded
0 atom has six valence electrons and se
ve
n associated
electrons (six in three lone pairs and one from the single bond). The doubly bonded
0 atom has six
valence electrons and six associated electrons (four in two lone pairs and two from the double bond).
Solution The formal charges are as follows: + 1 (N atom),

-1
(singly bonded 0 atoms), and 0
(doubly bonded 0 atom).
Practice Problem A Determine the formal charges on each atom in the carbonate ion
(
CO
~
-
)
.
Practice Problem B Determine the formal charges and use them to determine the overall charge,
if
any, on the species represented by the following Lewis structur
e:

I

:O-S-O:
• • • • • •
~I
~
________________________________________________________
~

While
you
are new
at
determ
ini

ng
fo
rmal
charges,
it
may
be
helpful to draw L
ewis
.
structures
with
all
dots,
rathe
r than d
ashes.
T
his
can
make
it
easier
to
see
how
ma
ny
electrons
ar

e
associated
with
ea
ch atom.
• •• • •
0:
O'
0:
.

Remember
that f
or
the
pu
r
pose
of counting
associated
elect
ron
s,
tho
se
s
har
ed
by
two at

oms
are
evenly
split between them.
Think
About
It
The sum
of
formal
charges
(+
1) + (
-1
) +
(-
1) +
(0) =
-1
is equal to the ov
er
all
charge on the nitrate ion.
294 CHAPTER 8 Chemical
Bonding
I:
Basic Concepts
Think
About
It

For a molecule,
formal charges
of
zero are
preferred. When there are
nonzero formal charges, they
should be consistent with the
electronegativities
of
the atoms
in
the molecule. A positive formal
charge on oxygen, for example, is
inconsistent with oxygen's high
electronegativity.
Sometimes,
there
is
more
than
one
possible
skeletal
arrangement
of
atoms
for
the
Lewis
structure

for
a
given
s
pecies.
In
such
cases,
we
often
can
select
the
best
skeletal
arrangement
by
using
formal
charges
and
the
following
guidelines:
•
For
molecule
s, a
Lewis
structure

in
which
all
formal
charges
are
zero
is
preferred
to
one
in
which
there
are
nonzero
formal
charges.
•
Lewis
structures
with
small
formal
charges
(0
and
+ 1)
are
preferred

to
those
with
large
for-
mal
charges
(+2,
+3,
and
so
on).
•
The
best
skeletal
arrangement
of
atoms
will
give
rise
to
Lewis
structures
in
which
the
formal
charges

are
consistent
with
electronegativities.
For
example,
the
more
electronegative
atoms
should
have
the
more
negative
formal
charges.
Sample
Problem
8.8
shows
how
formal
charge
can
be
used
to
determine
the

best
skeletal
arrangement
of
atoms
for
the
Lewis
structure
of
a
molecule
or
polyatomic
ion.
Sample Problem
8.8
.
Formaldehyde (CH
2
0),
which can be used
to
preserve biological specimens, is commonly sold as a
37% aqueous solution.
Use formal charges
to
detellTline which skeletal arrangement
of
atoms shown

here is the best choice for the Lewis structure
of
CH
2
0 .
H-C
-
O-H
o
I
H-C-H
Strategy
Complete the Lewis structures for each
of
the CH
2
0 skeletons shown and determine the
formal charges on the atoms in each one.
Setup The completed Lewis structures for the skeletons shown are:
• • • •
H-C=O-H
• •
'0'
II
H-C-H
In
the structure on the left, the formal charges are as follows:
Both H atoms: 1 valence
e - - 1 associated e - (from single bond) = 0
C atom: 4 valence e- - 5 associated e- (two in the lone pair, one from the single bond, and two from

the double bond)
= - 1
o ato
m:
6 valence e - - 5 associated e - (two from the lone pair, one from the single bond, and two
from the double bond)
= + 1
• • • •
H-C=O-H
Formal charges o - J + 1 0
In the structure on the right, the fOlmal charges are as follows:
Both H atoms: 1 valence
e - - 1 associated e - (from single bond) = 0
C atom: 4
va
lence e - - 4 associated e - (one from each single bond, and two from the double
bond)
= 0
o atom: 6 valence e - - 6 associated e - (four from the two lone pairs and two from the double
bond)
= 0
• •
'0'
II
H-C-H
Formal charges all zero
Solution
Of
the two possible arrangements, the structure on the left has
an

0 atom with a positive
formal charge, which is inconsistent with oxygen's high electronegativity. Therefore, the structure
on the right, in which both H atoms are attached directly
to
the C atom and all atoms have a formal
charge
of
zero, is the better choice for the Lewis structure
of
CH
2
0.
Practice Problem A Two possible arrangements are shown for the Lewis structure
of
a carboxyl
group,
-COOH.
Use formal charges to determine which
of
the two arrangements is better.
• •
'0'
II

• • • • • •
-C-O-H
-C-O=O-H


Practice Problem B Use Lewis structures and formal charges to determine the best skeletal

arrangement
of
atoms
in
NCI
2
.
Checkpoint 8.6 Lewis Structures and Formal Charge
8.6.1
Detennine the formal charges on H, C,
and
N,
respectively, in
HCN.
a) 0, +
1,
and - 1
b)
- 1, +
1,
and 0
c) 0, - 1, and + 1
d)
O,
+
I,and+l
e)
0,0,
and 0
Resonance

8.6.2 Which
of
the Lewis structures shown is
most likely preferred for
NCO
-?
a)
[~=C=qj
b)
~B-c=oJ
c)
~N=C-QJ
d)
~B-c
=q
J
e)
[~
=C-QJ
Our drawing
of
the Lewis structure for ozone
(0
3
)
satisfied the octet rule for the central 0 atom
because we placed a double bond between it and one
of
the two terminal 0 atoms. In fact, we can
put the double bond at either end

of
the molecule, as shown by the following two equivalent Lewis
structures:
'0=0-0:

<
-_.
'.

• • • • •
:0-0=0:

.
A single bond between 0 atoms should be longer than a double bond between 0 atoms, but experi-
mental evidence indicates that both
of
the bonds in 0
3
are equal
in
length (128 pm). Because
neither one
of
these two Lewis structures accounts for the known bond lengths in 0
3
,
we use both
Lewis structures
to
represent the ozone molecule.

Each
of
the Lewis structures is called a resonance structure. A resonance structure is one
of
two or more Lewis structures for a single molecule that cannot be represented accurately by only
one Lewis structure. The double-headed arrow indicates that the structures shown are resonance
structures. Like
th
e medieval European traveler to Africa who described a rhinoceros as a cross
between a griffin and a unicorn (two familiar but imaginary animals), we describe ozone, a real
molecule, in terms
of
two familiar but nonexistent structures.
A common
mi
sconception about resonance is that a molecule such as ozone somehow shifts
quickly back and forth from one resonance structure to the other. Neither resonance structure,
though, adequately represents the actual molecule, which has its own unique, stable structure.
"Resonance" is a human invention, designed to address the limitations
of
a simple bonding model.
To
extend the animal analogy, a rhinoceros is a distinct, real creature, not some oscillation between
the mythical griffin and unicorn!
The carbonate ion provides another example
of
resonance:
2-
?-
-

2-

'0'
• •

'0'
• •
'0'
• •
. I

'O=C-O:
• • • •
. . I .
:O-C=O
• • • •

II

:O-C
-O
:
• • • •
<
<
•
•
Ac
cording to experimental evidence, all three carbon-oxygen bonds in
CO

~-
are equivale
nt.
Therefore, the properties
of
the carbonate ion are best explained by considering its resonance
structures together.
The concept
of
resonance applies equally well to organic systems. A good example
is
the
benzene molecule (C
6
H
6
):
<
•
If
one
of
these resonance structures conesponded to the actual structure
of
benzene, there would
be two different bond lengths between adjacent C atoms, one with the properties
of
a single bond
and the other with the properties
of

a double bond. In fact, the
di
stan
ce
between all adjacent C
atoms in benzene
is
140
pm
, which is shorter than a
C-C
bond
(lS4
pm) and longer than a
C=C
bond (133 pm).
SECTION 8.7
Resonance
295
296 CHAPTER 8 Chemica l
Bonding
I:
Basic Concepts
Th
e
representati
on of
organic
compounds
is

d
is
cussed
in
mo
re
de
ta
il
in
Chapter 1
0.
Think
About
It
Always make sure
that resonance structures differ
only in the positions
of
the
electrons,
not
in the positions
of
the
atoms,
A simpler way
of
drawing the structure
of

the benzene molecule and other compounds con-
taining the benzene ring is to show only the skeleton and not the carbon and hydrogen atoms, By
this convention, the resonance structures are represented by
Note that the C atoms at the corners
of
the hexagon and the H atoms are not shown, although they
,
""
are' understood
to
'be'
th
'ere, Only the bonds between the C atoms are shown,
Resonance structures differ only in the positions
of
their electrons not in the positions
of
••
••
their atom'
s,
Thus,
:N=N=O:
and
:N-N
-
O:
are resonance structures
of
each other, whereas

••
••
:N=N=O:
and
:'N=O=N:
are not.
o • • •
Sample Problem 8,9 shows how to draw resonance structures,
High oil and gasoline prices have renewed interest
in
alternative methods
of
producing energy,
including the
"clean" burning
of
coal. Part
of
what makes "dirty" coal dirty is its high sulfur content.
Burning dirty coal produces sulfur dioxide
(S0
2)' among other pollutants, Sulfur dipxide
is
oxidized
in
the atmosphere to form sulfur trioxide
(S0
3), willch subsequently combines with water to produce
sulfuric
acid-a

major component
of
acid rain, Draw a
ll
possible resonance structures
of
sulfur trioxide,
Strategy
Draw two
or
more Lewis structures for
S0
3 in which the atoms are arranged the same way
but the electrons are arranged differently,
Setup Following the steps for drawing Lewis structures, we determine that a correct Lewis structure
for
S0
3 contains two sulfur-oxygen single bonds and one sulfur-oxygen double bond,

'0'
, ,
, I

:O=S-O:
,

But the double bond can be put in
anyone
of
three positions

in
the molecule,

, ,
• •
'0'
, .
U
'0'
, .
•
I

II

I
•
Solution
:O=S-O:
• •
:O-S
-
O:
• •
:O-S=O:
,





,
Practice Problem A Draw all possible resonance structures for the nitrate ion (NO) ,
Practice Problem B Draw three resonance structures for the tillocyanate ion (NCS- ), and determine
the formal charges in each resonance structure,
.
Checkpoint
8.7
Resonance
8.7.1
Indicate which
of
the following are 8.7.2
•• ••
resonarlce structures
of
:Cl
-Be-Cl:
• • • •
(s
elect all that apply),
, .
a)
:Cl=Be=Cr
• •

b)
:Cl=Be-Cl
:



c)
:Be=Cl
-
<:;:}:

d)
:CI-Be=Cl:

. ,
e)
:Be=C
l
=Cr
, .
Exceptions
to
the Octet Rule
How many resonance structures can
be drawn for the nitrite ion
(NO
z
)?
(N and 0 must obey the octet rule,)
a) 1
b) 2
c) 3
d) 4
e) 5
The octet rule almost always holds for second-period elements, Exceptions to the octet rule fall
into three categories:

SECTION
8.8 Exceptions
to
the
Octet Rule 297
I. The central atom has fewer than eight electrons due
to
a shortage
of
electrons.
2.
The central atom has fewer than eight electrons due
to
an odd number
of
electrons.
3.
The central atom has more than eight electrons.
Incomplete Octets
In so
me
compounds the number
of
electrons surrounding the central atom in a stable molecule
is fewer than eight. Beryllium, for example, which is the Group
2A
element in the second
period, has the electron configuration
[He]2i.
Thus, it has two valence electrons in the 2s

orbital. In the gas phase, beryllium hydride (BeH
2
)
exists as discrete molecules.
The
Lewis
structure
of
BeH
2
is
H-Be-H
Only four electrons surround the
Be
atom, so there is no way to satisfy the octet rule for beryllium
in this molecule.
Elements in Group 3A also tend to form compounds in which they are surrounded by fewer
than eight electrons. Boron, for example, has the electron configuration
[He]2
i2pi,
so it has only
three valence electrons. Boron reacts with the halogens to form a class
of
compounds having the
general formula BX
3
, where X is a halogen atom. Thus, there are only six electrons around the
boron atom in boron trifluoride:
• •
'F'

• •
· . I

:F-B-F:
• • • •
We
actually can satisfy the octet rule for boron in
BF
3 by using a lone pair on one of the F
atoms to form a double bond between the F atom and boron. This gives rise
to
three additional
resonance structures:
• •
:F:
. . I

F=B-F:
• • • •
• •
• •
.
F'
· .
II

:F-B-F:
• • • •
••
'F'

• •
. . I .
•
•
:F-B=F:
• •
•
Although these resonance structures result in boron carrying a negative formal charge while ftuo-
rine carries a positive formal charge, a situation that is inconsistent with the electronegativities of
the atoms involved, the experimentally determined bond length in
BF
3 (130
.9
pm) is shorter than a
single bond (137.3 pm). The shorter bond length would appear to support the idea behind the three
resonance structures.
On the other hand, boron trifluoride combines with ammonia in a reaction that is better rep-
resented using the Lewis structure in which boron has only six valence electrons around it:
• •
·F·
• •
. . I
:F
-B
+
. . I
:F:
• •
H
I

:N-H
I
H
••
:F:
H
. . I I
)
:F-B-N-H
. . I I
:F:
H
• •
It seems, then, that the properties
of
BF
3 are best explained by all four resonance structure
s.
The
B-N
bond in F
3
B-NH
3
is different from the covalent bonds discussed so far in the
ense that both electrons are contributed by the N atom. This type of bond is called a
coordinate
co
valent
bond

(also referred
to
as a dative bond), which is defined
as
a covalent bond in which
one
of
the atoms donates both electrons. Although the properties
of
a coordinate covalent bond do
not differ from those
of
a normal covalent bond (i.e., the electrons are shared in both cases), the
distinction is useful for keeping track
of
valence electrons and assigning formal charges.
Odd Numbers
of
Electrons
Some molecules, such
as
nitrogen dioxide (NO?), contain an odd number
of
electrons.
:O=N-O:
• • •
Because we need
an
even number
of

electrons for every atom in a molecule to have a complete
octet, the octet rule cannot be obeyed for all the atoms in these molecules. Molecules with an odd
~u
mber
of
electrons are sometimes referred to
as
free radicals (or
ju
st radicals). Many radicals
:rr
e highly reactive, because there is a tendency for the unpaired electron to form a covalent bond
.
'ith an unpaired electron on another molecule. When two nitrogen dioxide molecules collide, for
•
298
CHAPTER
8 Chemical Bonding
I:
Basic Concepts
The American Media
Inc.
building
in
Boca
Raton, Florida.
Severe flooding
in
New
Orleans after

Hurricane Katrina
in
2005.
Think
About
It C
l0
2
is used
primarily
to
bleach wood pulp
in
the manufacture of paper, but it is
also used to bleach flour,
di
sinfect
drinking water, and deodorize
certain industrial facilities.
Recently, it has been used to
eradicate the toxic mold in homes
in
New Orleans that were damaged
by the devastating floodwaters of
Hunicane Katrina in 2005.
The
OH
species
is
a

radical,
not to
be
confused
with
the
hydroxide
ion (OW).
example,
they
form
dinitrogen
tetroxide,
a
molecule
in
which
the
octet
rule
is satisfied
for
bot
h
the
Nand
0
atoms.
"
'l)".

,
,~
\
I '
N,
+
,N
, j
\
'0:
:0,
, '
••
Bringing Chemistry
to
Life
T
he
Power
of
Radica
ls
"
'l>:
:0
\\
I '
•
N-N
. 1

\\
:0.
0"
• •
••
Beginning
about
a
week
after
the
September
11, 2001, attacks,
letter
s
containing
anthrax
bacteria
were
mailed
to several
news
media
offices
and
to
two
U.S.
senators.
Of

the
22
people
who
subse-
quently
contracted
anthrax,
five died.
Anthrax
is a
spo
re-forming
b
ac
terium
(Bacillus anthracis)
and,
like
s
mallpox,
is
cla
ssified
by
the
CDC
as a Category A
biot
errorism agent.

Spore
-
forming
bacteria
are
notoriou
s
ly
difficult to kill,
making
the
cleanup
of
the
buildings
contaminated
by
anthrax
costly
and
time-con
s
uming.
The
American
Media
Inc.
(AMI)
building
in

Boca
Raton,
Florida
,
was
not
deemed
sa
fe
to
enter
until
July
of
2004,
after
it
had
been
treated
with
ch
l
orine
•
dioxide
(Cl
O?),
the
only

structural
fumigant
approved
by
the
Environmenta
l
Protection
Agency
(EPA)
for
anthrax
decontamination.
The
effectiveness
of
CI0
2
in
killing
anthrax
and
other
hardy
biological
agents
stems
in
part
from

it
s
being
a radical,
meaning
that
it
contains
an
odd
number
of
electrons.
Sample
Problem
8.10 .
Draw the Lewis stmcture
of
chlorine dioxide
(C
l0
2
) .
Strategy
The skeletal stmcture is
O-CI-O
This puts the unique atom, Cl, in the center and puts the more electronegative 0 atoms
in
terminal
pO

SItIO
n
s.
Setup
There are a total
of
19
valence electrons (seven from the Cl and s
ix
from each of the two 0
atoms).
We
subtract four electrons to account for the two bonds in the skeleton, leaving
us
with
15
electrons to distribute
as
follows: three lone pairs
on
each 0 atom, one lone pair on the
Cl
atom, and
the last remaining electron also on the
Cl
atom.
• • • • • •
Solution
:O-CI-O:
• • •

••
. . . . . . . . . .

. ,
Practice Problem A Draw the Lewis stmcture for the OH species.
Practice Problem B Draw the Lewis structure for the
NS
2
molecule.
Expanded Octets
Atoms
of
the
s
econd-period
elements
cannot
have
more
than
eight
valence
el
ectrons
around
them,
but
atoms
of
element

s in
and
beyond
the
third
period
of
the
periodic
table
can. In
addition
to
the
3s
and
3p orbitals,
elements
in
the
third
period
also
have
3d
orbita
ls
that
can
be

used
in
bo
ndi
ng
.
The
se
orbitals
enable
an
atom
to
form
an
expanded octet.
One
compound
in
which
there
is
an
expanded
octet
is
sulfur
hexafluoride
, a very
stable

compound.
The
electron
configuration
of
sul-
fur
is
[Ne]3i3p
4.
In
SF
6
,
each
of
sulfur's
6
valence
electrons
forms
a
covalent
bond
with
a fluorine
atom,
so
there
are

12
electrons
around
the
central
sulfur
atom:

:F:
"p'
I
F:
'.
"S


__

.0
:F
I F:
•• • •
:F:
••
SECTION
8.8 Exceptions
to
the
Octet
Rule 299

In Chapter 9 we will see that these
12
electrons, or six bonding pairs, are accommodated in
six orbitals that originate from the one
3s,
the three 3p, and two
of
the five
3d
orbitals. Sulfur also
forms many compounds in which it does obey the octet rule.
In
sulfur dichloride, for instance, S is
surrounded by only eight electrons:
I
• • • • • •
:Cl
-S
-Cl:
• • • • • •
Sample Problems 8.11 and 8.12 involve compounds that do not obey the octet rule.
.

_.
. .
Sample Problem 8.11. .
Draw the Lewis structure
of
boron triiodide
(El

3
).
Strategy
Follow the step-by-step procedure for drawing Lewis structures. The skeletal structure is
I
I
I-B-I
Setup There are a total
of
24 valence electrons in
BI
3 (three from the B and seven from each
of
the three I atoms).
We
subtract 6 electrons
to
account for the three bonds in the skeleton, leaving 18
electrons to distri bute as three lone pairs on each I atom.
Solution

. I .
• •

I

:I-B-I:
••
• •


. I .
• •

I

:I-B-I:


. I .
• •
•
I

• •
:I=B-I:
•

• •
Think
About
It
Boron is one
of
the elements that does not always
follow the octet
rule. Like
BF
3
,
however, BI3 can

be
drawn with a
double bond in order to satisfy the
octet
of
boron. This gives rise to a
total
of
four resonance structures:
• •

T
. I .
• •

II


I
•
:I-B-I:
•
•
:I-B=I:

•
• •
••
••••••••••••••••••••••••••••••
••••

•
Practice Problem A Draw the Lewis structure
of
beryllium fluoride (
BeF
2
).
Practice Problem B Draw the Lewis structure
of
boron trichloride (BCI
3
).
~I
__________________________________________________________
~

Draw the Lewis structure
of
arsenic pentafluoride (AsFs).
Strategy
Follow the steps for drawing Lewis structures. The skeletal structure already has more than
an octet around the As atom.
F
I / F
F-As
I " F
F
Setup There are 40 total valence electrons [five from As (Group
SA
) and seven from each

of
the five
F atoms (Group 7 A)]. We subtract
10
electrons to account for the five bonds in the skeleton, leaving
30 to be distributed. Next, place three lone pairs on each F atom, thereby completing all their octets
and using
up
all the electrons.
Solution

:F:
.

I
;f:
:F-As

I "
Ii:
'F'
" .
• •

Practice Problem A Draw the Lewis structure
of
phosphorus pentachloride (PCI
s
).
Practice Problem B Draw the Lewis structure

of
antimony pentafluoride (
SbF
s
)'


8eF
2
is
also
known
as
beryllium
difluoride, but
because
this
is
the
only
compound
that
beryllium
forms with fluori
ne,
t
he
name
beryllium fluoride
is

unambiguous.
Think
About
It
Always make
sure that the number
of
electrons
represented in your final Lewis
structure matches the total number
of
valence electrons you are
supposed to
have.
300 CHAPTER 8 Chemical
Bonding
I:
Basic
Concepts
Think
About
It
Atoms beyond the
second period can accommodate
more than an octet
of
electron
s,
whether those electrons are used
in bonds

or
reside on the atom as
lone pairs.
When drawing Lewis structures
of
compounds containing a central atom from the third
period and beyond, the octet rule may be satisfied for all the atoms before all the valence electrons
have been used up. When this happens, the extra electrons should
be
placed as lone pairs on the
central atom. Sample Problem 8.13 illustrates this approach.
Sample Problem 8.13
••
Draw the Lewis structure
of
xenon tetrafluoride (XeF
4
).
Strategy
Follow the steps for drawing Lewis structures. The skeletal structure is
F
I
F-Xe-F
I
F
Setup
There are 36 total valence electrons (eight from Xe and seven from each
of
the
fOllr

F atoms).
We subtract eight electrons to account for the bonds in the skeleton, leaving 28 to distribute. We first
complete the octets
of
all four F atoms. When this is done, four electrons remain, so we plilce two
lone pairs on the
Xe
atom.
Solution
• •
:F:
. . . . I

:F- Xe-
F:
. . I . . . .
:F:
• •
Practice Problem A Draw the Lewis structure
of
the iodine tetrachloride ion (ICI
4
).
Practice Problem B Draw the Lewis structure
of
krypton difluoride
(KrF
2)'
Checkpoint 8.8 Exceptions to the Octet Rule
8.8.1 In which

of
the following species does 8.8.3 In which species does the central atom
the central atom
not obey the octet obey the octet rule? (Select a
ll
that
rule? apply.)
a)
CI0
2
a)
I)
b)
CO
2
b)
BH
3
c)
BrO) c)
AsF6
d)
HeN
d)
N0
2
e)
ICl
4
e)

CI0
2
8.8.2
Which elements cannot have more than
8.8.4
How many lone pairs are there on the
an octet
of
electrons? (Select all that
central atom in the Lewis structure
of
apply.) ICI
2
?
a)
N a)
0
b)
C
b)
I
c)
S
c)
2
d)
Br
d)
3
e)

0
e) 4
Bond
Enthalpy
One measure
of
the stability
of
a molecule is its bond enthalpy, which is the enthalpy change asso- .
ciated with breaking a particular bond in 1 mole
of
gaseous molecules. (Bond enthalpies in solids
and liquids are affected by neighboring molecules.) The experimentally determined bond enthalpy
of
the diatomic hydrogen molecule, for example, is
H2(g) -
-+.
H(g)
+
H(g)
i1W
=
436.4
kJ/mol
SECTION
8.9 Bond Enthalpy
301
According to this equation, breaking the covalent
bond
s in 1 mole

of
ga
seous H2 molecules
requires 436.4 kJ
of
energy.
For
the less stable chlorine molecule,
Cl
i g)
+.
CI(g) + CI(g)
D.W
= 242.7 kJ/mol
Bond
enthalpies can also
be
directly
mea
sured for heteronuclear diatomic molecules, s
uch
as
HCI,
HCI(g)
+.
H(g) + Cl(g)
as well as for molecules containing multiple bonds:
0 2
(g
)

+.
O(g) + O(g)
N
2
(g) • N(g) +
N(
g)
D.W
= 431.9 kJ/
mol
D.W
= 498.7 kJ/
mol
D.W
=
941.4kJ
/
mol
Measuring the strength
of
covalent bonds in poly atomic molecules is
more
complicated.
For
example, measurements show that the energy needed to break the first
0-
H bond in H?O is differ-
ent from that needed to
break
the second

0-
H bond:
H
2
0(g)
+.
H(g) +
OH(
g)
OH(g
) • H(g) + O
(g
)
D.W
= 502
kJ
/
mol
D.H
o = 427
kJ
/
mol
In
each case, an
0-
H bond is broken, but the first step requires the input
of
more
energy than the

second.
The
difference between the two
D.H
o values suggests that the second
0-
H bond itself
undergoes change, because
of
the changes in its chemical environment.
We can now understand why the bond enthalpy
of
the s
ame
0-
H bond in two different
molecules, such as methanol (CH
3
0H)
and water (H
2
0 ), will not be the same: their environments
are different.
For
polyatomic molecules, therefore, we speak
of
the average bond enthalpy
of
a particular bond.
For

example, we can measure the enthalpy
of
the
0-
H
bond
in 10 different
polyatomic molecules and obtain the average
0-
H bond enthalpy by dividing the s
um
of
the bond
enthalpies
by
10. Table 8.6 lists the average bond enthalpies
of
a number
of
diatomic and poly-
atomic molecules. As we noted earlier, triple bonds are stronger than double bonds, and double
bonds are stronger than single bonds.
A comparison
of
the thermochemical changes that take place during a
number
of
reactions
reveals a strikingly wide variation in the
en

thai pies
of
different reactions.
For
example, the com-
bustion
of
hydrogen gas in oxygen
ga
s is fairly exothermic:
D.H
o = - 285.8 kJ/
mol
The formation
of
glucose from carbon dioxide and water, on the other hand, best achiev
ed
by
pho-
tosynthesis, is highly
endothermic:
D.H
0 = 2801 kJ/mol
We can account for such variations
by
looking at the stability
of
individual reactant and product
molecules. After all, most chemical reactions involve the making and breaking of
bond

s. There-
fore, knowing the bond enthalpies and hence the stability
of
molecules
re
veals something about
rhe thermochemical nature
of
the reactions that molecules undergo.
In many cases, it is possible to predict the approximate enthalpy
of
a reaction by using the
av
erage
bond
enthalpies. Because energy is always required to
break
chemical
bond
s and chemical
bond formation is always accompanied
by
a release
of
energy, we can estimate the enthalpy
of
a
reaction
by
counting the total

number
of
bonds broken and formed in the reaction and recording all
the corresponding enthalpy changes.
The
enthalpy
of
reaction
in
the
ga
s
pha
se is given by
D.W
= 2,BE(reactants) - 2,BE(products) Equation 8.3
= total energy input (to
br
eak
bond
s) - total energy released (by
bond
formation)
. .
where
BE
stands for average bond enthalpy and
2,
is the summation sign. As written,
Equation

8.3
rakes care
of
the sign convention for
D.H
o. Thus,
if
the total energy input needed to
break
bond
s
in
the reactants is less than the total energy released when bonds are formed in the products, then
~
o
is negative and the reaction is exothermic [Figure 8.8(a)]. On the other hand, "ifless energy is
released (bond making) than absorbed (bond breaking),
D.H
o is
po
sitive and the reacrion is endo-
thermic [Figure 8.8(b)].
•
To
many
students,
Equation
8
.3
ap

p
ea
rs
to be
backward.
Ordinarily
you
calcu
l
ate
a
change
as
final
minus
initial.
Her
e
we
a
re
determining the
differe
nce
between
the
amount of
heat
we
ha

ve
to
add
to
break
reactant
bonds
and t
he
amount
of
heat
releas
ed when product
bon
ds form. The
sign
of
the
final
answ
er
tells
us
if t
he
p
ro
ces
s

is
endo
t
hermic
(+ ) or exothermic
(-)
o
ve
ra
l
l.
Bond
ent
halp
i
es
for diatomic
mol
e
cules
have
more
significant figures than
those
for
polyatom
ic
mo
l
ecules.

Those
for
poly
atomic
molec
ul
es
are
av
er
age
va
l
ues
base
d
on
the
bonds
in
more than
one
compound.
Figure 8.8 Enthalpy changes in
(a) an exothermic reaction and
(b
) an
endothermic reaction. The
t:.H
o values

are calculated using Equation 5.19 and
tabulated
t:.H
? values from Appendix 2.
302
• • •
Bond
H- H*
H-N
H-O
H-S
H-P
H- F
H -
Cl
H-
Br
H - I
C-H
c-c
c=c
c=c
C-
N
C=N
C- N
c- o
C=o
t
c-o

C-P
Bond
Enthalpy
(kJ/mol)
436.4
393
460
368
326
568.2
431.9
366.1
298.3
414
347
620
812
276
615
891
351
745
1070
263
•
•••
••
• • • •
••
•

* Bond enthalpies show n in red are
for
diatomic
molecules.
' The
c=o
bond
enthalp
y in
CO
2
is 799
kJ
/
mo
!.
Reactant
molecul
es
Atoms
Product
molecules
H
2
(g) + Cli g) • 2HCI(g)
t!.H
o = - 18
4.
7 kJ/
mol

(a)
Bond
c-s
c=s
N-N
N=N
N=N
N-O
N=O
0 - 0
0 =0
O-P
o=s
P-P
P=P
s-s
s=s
F- F
CI-
Cl
Cl
- F
Br
-
Br
I - I
Reactant
mole
c
ules

Atoms
Bond
Enthalpy
(kJ/mol)
255
477
193
418
941.4
176
607
142
498.7
502
469
197
489
268
352
156.9
242.7
193
192.5
151.0
Product
molecules
2NH
3
(g) • N
2

(g) + 3H
2
(g)
t!.H
O = 92.6
kJ
/
mol
(b)
SECTION
8.9 Bond Enthalpy 303
If
all the reactants and products are diatomic mol
ec
ule
s,
then the equation for the enthalpy
of
reaction will yield accurate results because the bond enthalpies
of
diatomic molecules are accu-
rately known.
If
some or all of the reactants and products are poly atomic molecules, the equation
will yield only approximate results because
th
e bond enthalpies used will be averages.
I
I
I

Sample Problem 8.14 shows how to estimate enthalpies
of
reaction us
in
g bond enthalpies.

Sample Problem 8.14


Use bond entha1pies from Table 8.6 to estimate the enthalpy
of
rea
ct
ion for the combustion
of
methane:
• •
Strategy
Draw
Lewis structures to determine what bonds are to be broken and what
bond
s are to
be
formed.
Setup
H
I
H-C-H
I
H

• •
:0=0:
• •
+
• •
:0=0:
• •
Bond
s to break: 4
C-H
and 2
0=0.
Bond
s to form: 2
C=O
and 4
H-O
.

H-O-H

• •
•
:O=
C
=O:
+
• •

H-O-H


Bond
enthalpies from Table 8.6:
414
kJ
/mol
(C-
H
),
498.7 kJ/mol
(0=0
),
799
kJ/m01
(C=
O in
CO
2
),
and
460
kJ
/mol (H
-0)
.
Solution
[4(414
kJ
/mol) + 2(498.7 kJ/mol)] - [2(799
kJ

/mol) + 4(460 kJ/mol)] = - 785 kJ/mo
!.
Practice
Problem
Use
bond enthalpies from Table 8.6 to estimate the enthalpy
of
reaction for the
combination
of
carbon monoxide
and
oxygen to produce carbon dioxide:
~

Bond Enthalpy
8.9.1
Us
e data from Table 8.6 to
est
imate
8.9.2
Use
data from Table 8.6 to estimate
t1H
rxn
for the reaction
of
ethylene with
t1H

r
xn
for the reaction
of
fl
uorine and
hydrogen to produce ethane.
chlorine to produce ClF.
C
2
H
4
(g) + H
2
(g)
• C
2
H
6
(g)
F
ig)
+ CI
2
(g)
•
2C
lF
(g)
a)

-1
19
kJ/mol
a)
-
77
.5
kJ
/mol
b)
11
9
kJ
/mol
b)
-206
.6
kJ/mol
c)
-392
kJ
/mol
c)
206.6
kJ
/mol
d) 392
kl
lmol
d)

-13.6
kJ/mol
e)
-828
kJ
/mol
e) 13.6
kJ
/mol
Don't
skip
the
ste
p of drawing L
ewis
struc
tures.
This
is
the
only way to know for
certain
what
types
and
numbers
of
bonds
must
be

broken
and
form
ed
.
Think About It Use Equation
5.19
[
~.
Section 5.6] and data from
Appendix 2
to
calculate this enthalpy
of
reaction again; then compare your
results using the two approaches.
The difference in this case is due
to two things: Most tabulated bond
enthalpies are averages and, by
convention, we show the product
of
combustion as liquid wa
ter-but
average bond enthalpies apply to
species
in the gas phase, where there
is little or no influence exerted by
. molecules.
Remember
that

heats
of
reaction
are
expressed
in
kJ/mol,
where the "per mole"
refers
to
pe
r
mole
of
reaction
as
written
[H
. SectIOn 5.3] .
304
CHAPTER
8 Chemical
Bonding
I:
Basic
Concepts
Applying
What
You've Learned
Researchers in the early 1990s made the sensational announcement that nitric oxide

(NO), which had long been thought
of
only as a component
of
air pollution, turns out to
play an important role in human physiology. They found that
NO serves
as
a signal mol-
ecule, being produced
in vivo and regulating a wide variety
of
cell functions in the body
including in the cardiovascular, nervous, and immune systems.
The discovery
of
the biological role
of
nitric oxide has shed light on how nitro-
glycerin works as a drug. For many years, nitroglycerin tablets have been prescribed for
cardiac patients to relieve the pain caused by brief interruptions in the flow
of
blood to
the heart, although how it worked was not understood. We now know that nitroglycerin
produces nitric oxide in the body, which causes muscles to relax and allows the arteries
to dilate.
Research continues to uncover the role nitric oxide plays in biological processes,
and medicine continues to find new uses for this molecule.
Problems:
a) Without consulting Figure 8.1, give the Lewis dot symbols for

Nand
O.
[
~~
Sample
Problem
8.1]
•
b) Classify the bond in NO
as
nonpolar, polar, or ionic.
[I
~~
Sample
Problem
8.4]
c) Given the experimentally determined dipole moment (0.16 D) and the bond length
o
(1.15 A), determine the magnitude
of
the partial charges in the NO molecule.
[
~~
Sample
Problem
8.5]
d)
Draw the Lewis structures for NO and for nitroglycerin (C3HsN30 9
)'
[

~~
Sample
Problem
8.6]
e) Determine the formal charges on each atom in
NO and in nitroglycerin.
[
~~
Sample
Problem
8.7]
f) Nitroglycerin decomposes explosively to give carbon dioxide, water, nitrogen, and
oxygen. Given the balanced equation for this reaction,
use bond enthalpies to estimate
D.H
o for the reaction.
[
~~
Sample
Problem
8.13]
CHAPTER SUMMARY
Section 8.1
•
A
Lewis
dot
symbol
depicts an atom
or

an atomic ion
of
a main group
element with dots (representing the valence electrons) arranged around '
the element's symbol. Main group atoms lose
or
gain one or more
electrons to become isoelectronic with noble gases,
Section 8.2
•
•
The electrostatic attraction that holds ions together in an ionic
compound is refen"ed to as
ionic
bonding.
Lattice
energy
is the amount
of
energy required to convert a mole
of
ionic solid to its constituent ions in the gas phase. Lattice energy
cannot
be
measured directly, but is determined using the
Born-Haber
cycle and thermodynamic quantities that can be
measmed
directly.
Section 8.3

•
•
According to the
Lewis
theory
of
bonding,
covalent
bonding
results
when atoms
share valence electrons.
The
atoms in molecules and
those
in
polyatomic ions are held together by
covalent
bonds.
According to the octet rule, atoms will lose, gain, or share electrons in
order to achieve a noble gas configuration. Pairs
of
valence electrons that
are
not involved in the covalent b0nding in a molecule or polyatomic ion
(i.e., valence electrons that are not shared) are called
lone pairs.
•
Lewis
structures

are drawn to represent molecules and polyatomic
ions, showing the arrangement
of
atoms and the positions
of
all
valence electrons. Lewis structures represent the shared pairs
of
valence electrons either as two dots, ,
or
as a single dash,

Any
unshared electrons are represented as dots.
•
One
shared pair
of
electrons between atoms constitutes a
single
bond.
Multiple
bonds
form between atoms that share
more
than one pair
of
electrons. Two shared pairs constitute a
double
bond,

and three shared
pairs constitute a
triple
bond.
Section 8.4
•
Bonds
in
which electrons are not shared equally are
polar
and are
referred to as
polar
covalent
bonds.
• Electronegativity is an
atom's
ability to draw shared electrons toward
itself. Bonds between elements
of
widely different electronegativities
(~
:>
2.0) are ionic. Covalent bonds between atoms with significantly
different electronegativities
(0.5 <
~
< 2.0) are
palm:
Bond

s between
atoms with very similar electronegativities
(~
< 0.5) are nonpolar.
KEyWORDS
Bond enthalpy,
300
Born-Haber cycle, 282
Coordinate covalent bond, 297
Covalent bond, 284
Covalent bonding, 284
Dative bond, 297
Dipole
moment
(f.L),
289
KEY EQUATIONS
Double bond, 285
Electronegativity, 287
Formal charge,
292
Free
radical, 297
Ionic bonding, 279
Lattice energy,
280
8.1
f.L=Q
X r
KEY EQUATIONS 305

• The dipole
moment
(p )
is a quantitative measme
of
the polarity
of
a bond.
Section 8.5
•
Lewis structures
of
molecules
or
polyatomic ions can
be
drawn using
the following step-by-step procedure:
1.
Use
the molecular formula to draw the skeletal structure.
2.
Count the total number
of
valence electrons, adding electrons to
account for a negative charge
and
subtracting electrons to account
for a positive charge.
3.

Subtract two electrons for each bond in the skeletal structure.
4.
Di
stribute the remaining valence electrons to complete octets,
completing the octets
of
the
more
electronegative atoms first.
5.
Place any remaining electrons on the central atom.
6.
Include double
or
triple bonds,
if
necessary, to complete the octets
of
all atoms.
Section 8.6
•
Formal
charge
is a way
of
keeping track
of
the valence electrons in a
species. Formal charges should
be

consistent with electronecrativities
'"
and can be used to determine the
be
st arrangement
of
atoms and
electrons for a Lewis structure.
Section 8.7
•
Resonance
structures
are two
or
more
equally correct Lewis
structures that differ in the positions
of
the electrons but not in the
positions
of
the atoms. Different resonance structures
of
a compound
can
be
separated
by
a resonance arrow, • •
Section 8.8

• In an ordinary covalent bond, each atom contributes one electron
to the shared pair
of
electrons. In cases where
just
one
of
the atoms
contributes
both
of
the electrons, the bond is called a
coordinate
covalent
bond
or
a dative
bond.
• A species that contains an odd number
of
electrons is called
afree
radical.
Section 8.9
•
Bond
enthalpy
is the energy required to break I mole
of
a particular

type
of
bond.
Bond
enthalpies are a measure
of
the stabiiity
of
covalent
bonds and can be us
ed
to estimate the enthalpy change for a reaction.
Lewis dot symbol, 278
Lewis structure, 284
Lewis theory
of
bonding, 284
Lone pair, 284
Multiple bond, 285
Nonpolar, 288
Octet rule, 284
Polar, 286
Polar covalent bond, 286
Resonance structure, 295
Single bond, 285
Triple bond, 285
8.2 formal charge
= valence electrons - associated electrons
8.3
~H

O
= 2:BE(reactants) - 2:BE(product
s)
306
CHAPTER
8 Chemical
Bonding
I:
Basic Concepts
,
QUESTIONS AND PROBLEMS '
§@Ui@R
§:1!
b@wi§
~€}t
8Vm8€}1§
Review Questions
8.1
8.2
What
is a Lewis dot symbol?
What
elements do
we
generally
represent with
Lewi
s
sym
bol

s?
Use the second
member
of
each
group from Group 1A to Group
7 A to show that the
number
of
va
lence
electrons on an
atom
of
the
element
is the
same
as
it
s group numbe
r.
8.3 Without refen'ing to Figure 8.1 , write Lewis dot
sym
bols for
atoms
of
the following elements: (a) Be, (b) K, (c) Ca, (d) Ga,
(e)
0 ,

(f)
Br, (g)
N,
(h) I, (i) As, (j)
F.
8.4
Write
Lewis dot symbols for the following ion
s:
(a)
Li
+, (b)
CI-,
(c)
S2-,
(d)
Sr
2
+,
(e) N
3

8.5
Write
Lewis
dot
symbols for the following atoms and ions: (a) I,
(b)
r-, (c) S, (d)
S2-,

(e) P,
(f)
p
3
-,
(g) Na, (h) Na+, (i)
Mg,
(j)
Mg
2+,
(k) AI, (1) As
3+
, (m) Pb, (n) Pb
2+
.
§@€tieR
§:~llefli€
OeRsil1§
Review Questions
8.6 Explain
what
ionic bonding is.
8.7
8.8
8.9
8.10
8.11
8.12
8.13
8.14

8.15
8.16
8.17
Explain how ionization energy and electron affinity determine
whether
atoms
of
elements will combine to form ionic
compounds.
Name
five metals and five nonmetals that are very likely to
form
ionic compounds.
Write
fonnulas
for
compound
s that might
result from the combination
of
these metals and nonmetals.
Name
these compounds.
Name
one ionic compound that contains only nonmetallic elements.
Name
one
ionic
compound
that contains a

pol
y atomic cation and
a polyatomic anion (see Table 2.9).
Explain why ions with charges
greater
than
+3
are se
ldom
found
in ionic compounds.
The
term molar
mass
was introduced in
Chapter
3.
Molar mass
is numerically equivalent to molecular
ma
ss, although the units
are different, for a covalent compound.
What
is the advantage
of
using the
tenn
molar
mass when
we

discuss ionic compounds?
In
which
of
the following states would NaCI
be
electrically
conducting? (a) solid, (b) molten (that is, melted), (c) dissolved
in
water. Explain.
Beryllium
forms a
compound
with chlorine that has the empirical
formula BeCI
2
.
How
would you determine
whether
it is an ionic
compound?
(The
compound
is not soluble in water.)
What
is lattice energy, and
what
does it indicate
about

the
stability
of
an ionic
compound?
Explain how the lattice energy
of
an ionic
compound
such as KCl
can
be
determined using the
Born-Haber
cycle.
On
what law is
this procedure
ba
sed?
Specify which
compound
in
each
of
the following pairs
of
ionic
compound
s should

ha
ve the
higher
lattice energy: (a) KCI or
MgO,
(b)
LiF
or
LiBr
, (c) Mg3N2 or NaC!. Explain your choice.
8.18
Specify
which
compound
in
each
of
the following pairs
of
ionic
compounds should have the higher lattice energy: (a)
A1N
or
CaO, (b)
NaF
or
CsF, (c)
MgCl
2
or

MgF
2
. Explain your choice.
Problems
8.19
8.20
8.21
8.22
8.23
8.24
An
ionic
bond
is formed between a cation A + and an anion B

Ba
sed on
Coulomb's
law
how
would the energy
of
the ionic
bond
be
affected by the
following changes? (a) doubling the radius
of
A +, (b) tripling
the charge on A

+,
(c) doubling the charges on A + and B
-,
(d) decreasing the radii
of
A + and B - to
half
their original values.
Give the empirical formulas and names
of
the
compounds
formed
from
the
following pairs
of
ions: (a) Rb+ and
I-,
(b)
Cs
+ and
SO~-,
(c)
Sr
2
+
andN
3
- , (d)

AI
3+
and S2
Use
Lewis dot symbols to show the transfer
of
electrons between
the following atoms to
form
cations and anions: (a)
Na
and
F,
(b) K and S, (c)
Ba
and
0,
(d) Al and N.
Write the Lewis dot symbols
of
the reactants and products in the
following reactions. (First balance the equations.)
(a)
Sr
+
Se
-_.
SrSe
(b)
Ca

+ H2 •
CaH
2
(c)
Li
+ N
z
-_.
Li3N
(d) Al
+ S • AI
2
S
3
Use
the
Born-Haber
cycle outlined in Section 8.2 for
NaCl
to
calculate the lattice energy
of
LiC!.
Use
data from Figures 7.8
and
7.10
and Appendix 2.
Calculate the lattice energy
of

CaCl
z
.
Use data from Figures 7.8
and
7.10
and Appendix 2. (The second ionization energy
of
Ca,
lE
z
,
is 1145 kllmo!.)
8@etisflll31
EeVai@nt
Oenaiflg
Review Questions
8.25 Describe
Lewis's
contribution to our understanding
of
the
covalent bond.
8.26
Use
an
example
to illustrate
each
of

the following terms: lone
pair, Lewis structure,
the octet rule,
bond
length.
8.27
What
is the difference between a Lewis symbol and a Lewis
structure?
8.28
8.29
8.30
8.31
How many lone pairs are on the underlined atoms in these
compounds: HBr,
H:z.S.
,
CH
4
?
Compare
single, double, and triple bonds in a molecule, and give
an example
of
each.
For
the same
bonding
atoms,
how

does the
bond
length change
from
single
bond
to triple bond?
Compare
the properties
of
ionic
compounds
and covalent
compounds.
Summarize
the essential features
of
the Lewis octet rule.
The
octet rule applies mainly to the
second
period
elements. Explain.
Problems
8.32
For
each
of
the following pairs
of

elements, state
whether
the
binary
compound
they
form
is likely to
be
ionic
or
covalent.
8.33
Write the empirical formula and name
of
the compound:
(a) I and CI, (b)
Mg
and
F.
For each
of
the following pairs
of
elements,
st
ate whether the
binary compound they form is likely to be ionic or covalent.
Write the empirical formula and name
of

the compound:
(a)
Band
F,
(b) K and Br.
Section 8.4: Electronegativity and Polarity
Review Questions
8.34
8.35
Define electronegativity, and explain the difference between
electronegativity and electron affinit
y.
Describe in general how
the electronegativities
of
the elements change according to their
position in the periodic table.
What
is a
polar
covalent
bond
?
Name
two compounds that
contain one
or
more polar covalent bonds.
Problems
8.36 List the following bonds in order

of
increasing ionic character:
the lithium-to-fluorine bond in LiF, the potassium-to-oxygen
bond in
K
2
0,
the nitrogen-to-nitrogen bond in N
2
,
the sulfur-to-
oxygen bond in
S0
2,
the chlorine-to-fluorine bond in CIF
3
.
8.37 An-ange the following bonds in order
of
increasing ionic
character: carbon to hydrogen, fluorine to hydrogen, bromine to
hydrogen, sodium to chlorine, potassium to fluorine, lithium to
chlorine.
8.38
8.39
8.40
Four atoms are arbitrarily laheled D, E, F, and G. Their
electronegativities are as follows: D
= 3.8, E = 3.3, F = 2.8,
and G

= 1.3.
If
the atoms
of
these elements form the molecules
DE
,
DG
, EG, and DF, how would you an-ange these molecules in
order
of
increasing covalent bond character?
List the following bonds
in
order
of
increasing ionic character:
cesium to fluorine, chlorine to chlorine, bromine to chlorine,
silicon to carbon.
Classify the following bonds as ionic, polar covalent, or nonpolar
covalent, and explain: (a) the
CC
bond in H
3
CCH
3
,
(b) the KI
bond in KI, (c) the NB bond in H
3

NBCI
3
, (d) the
CF
bond in
CF
4
,
8.41
Cla
ssify the following bonds as ionic, polar covalent, or nonpolar
covalent, and explain: (a) the
SiSi bond in
CI
3
SiSiCl
3
,
(b) the
SiCl bond in Cl
3
SiSiCI
3
,
(c) the
CaF
bond in
CaF
2
,

(d) the
NH
bond in
NH
3
.
Section 8.5:
Drawing
lewis
Structures
Problems
.42
.43
.44
.45
Draw Lewis structures for the following molecules and
ions: (a) NCI
3
,
(b) OCS, (c) H
2
0
2
,
(d) CH
3
COO
- , (e)
CW,
(f)

CH3CH2NH; .
Draw Lewis structures for the following molecules and ion
s:
(a) OFz, (b) N
2
F
2
,
(c) Si
2
H
6
,
(d)
OW,
(e)
CH
2
CICOO- ,
(f)
CH3NH; .
Draw Lewis structures for the following molecules: (a) ICl,
(b)
PH
3
,
(c) P
4
(each P is bonded to three other P atoms), (d) H
2

S,
(e) N
2
H
4
,
(f)
HC10
3
, (g)
COBr2 (C is bonded to 0 and
Br
atoms).
Draw Lewis structures for the following molecules: (a)
ClF
3
,
(b) H
2
Se, (c)
NH
2
0H
, (d) POCl
3
(P is bonded to 0 and Cl
atoms), (e)
CH
3
CH

2
Br,
(f) NCI
3
,
(g)
CH
3
NH
2
·
QUESTIONS
AND
PROBLEMS 307
Section 8.6:
lewis
Structures and Formal Charge
Review Questions
8.46 Explain the concept
of
Jar
mal
charge. Do formal charges
represent
an
actual separation
of
charges?
Problems
8.47

8.48
8.49
8.50
Dr
aw Lewis structures for the following ions: (a)
NO~,
(b)
SCN
- , (c) s
i-,
(d)
CIF
~
.
Show formal charges.
Draw Lewis structures for the following ions: (a)
oi-, (b) ci-,
(c)
NO
+,
(d)
NH
! . Show formal charges.
The
skeletal structure
of
acetic acid shown here is con-ect, but
some
of
the bonds are wrong. (a) Identify the incorrect bonds

and explain what is wrong with them. (b) Write the correct Lewis
structure for acetic acid.
H
:0:
J J
H=C-C-O-H
J . . . .
H
The following Lewis structures are incorrect. Explain what is
wrong with each one, and give a
con-ect Lewis structure for the
molecule. (Relative positions
of
atoms are shown correctly.)
• • • •
(a)
H-C=N
••
(b)
H=C=C=H
• • • •
(c)
O-Sn-O
• • • •
••
• •
:F F:
(d) "
"B/
"

J
:p:
••
• • • •
(e)
H-O=!::
H
\


(f)
C-F:
. /

'0·
•
• • • •
:K
/F:
.0

N/
.0
(g) J
:p:
••
Section 8.7: Resonance
Review Questions
8.51
8.52

What
is a resonance
st
ructure? Is it possible to isolate one
resonance structure
of
a compound for analysis? Explain.
What
are the rules for writing resonance structures?
Problems
8.53 Write Lewis structures for the following specie
s,
including
all resonance forms, and show formal charges: (a)
HCO
2,
8.54
8.55
8.56
(b)
CH
2
N0
2
.
The
relative positions
of
the atoms are as follows:
o H o

H C C N
o H o
Draw three resonance
st
ructures for the chlorate ion (CIO
"3).
Show formal charges.
Write three resonance structures for hydrazoic acid (HN3)' The
atomic
an-angement is HNNN. Show formal charges.
Draw two resonance structures for diazomethane (CH
2
N
2
).
Show
formal charges. The skeletal structure
of
the molecule is
H
C N N
H
308
CHAPTER
8 Chemical
Bonding
I:
Basic Concepts
/
~'

8.57
8.58
8.59
Draw
three reasonable resonance structures for the
OCN-
ion.
Show
fonnal
charges.
Draw
three resonance structures for the molecule N
2
0 in which
the atoms are arranged in the order
NNO. Indicate
fonnal
charges.
Draw
a resonance structure
of
the adenine molecule shown here,
which is
part
of
the
DNA structure:
N
H
·NH

2
/
N
s@€liOfl
o:oi
E*€@~tl@R§
t@
tfl@
~€t@t
ftl:ll@
Review Questions
8.60
Why
does the octet rule not
hold
for many compounds containing
elements in the third period
of
the periodic table and beyond?
8.61 Give three examples
of
compounds that do not satisfy
the
octet
rule. Write a Lewis structure for each.
8.62 Because fluorine
ha
s seven valence electrons
(2i2
p

5),
seven
covalent bonds
in
principle
could
form around the atom. Such a
compound
might
be
FH7
or
FCI
7
.
These
compounds
ha
ve
ne
ver
been
prepared.
Why?
8.63
What
is a coordinate covalent bond? Is it different from an
ordinary covalent
bond?
Problems

8.64
8.65
8.66
8.67
8.68
8.69
The
AII3
molecule has an incomplete octet around Al.
Draw
three
resonance
structures
of
the
molecule
in which the octet rule is
satisfied for both the Al and the I
atom
s. Show formal charges.
In
the vapor phase, beryllium chloride consists
of
discrete
BeCl
2
molecules. Is the octet rule satisfied for
Be
in this compound?
If

not, can you form an octet around
Be
by drawing another
resonance structure? How plausible is this structure?
Of
the
noble gases, only Kr, Xe,
and
Rn
are known to form a
few compounds with
° andlor
F.
Write
Lewis structures for the
following molecule
s:
(a)
XeF
z
,
(b)
XeF
4
,
(c)
XeF
6
,
(d)

XeOF
4
,
(e)
XeO
zF
z
.
In
each
case
Xe
is the central atom.
Write a Lewis structure for
SbCl
s
.
Does this molecule obey the
octet rule?
Write
Lewis structures for SeF4
and
SeF
6
.
Is the octet rule
satisfied for
Se?
Write Lewis structures for the reaction
AICl

3
+
CI-
-_.
AICI
4
What
kind
of
bond
joins
Al and CI in the product?
Review Questions
8.70
8.71
What
is bond enthalpy?
Bond
enthalpies
of
polyatomic molecules
are average values, whereas those
of
diatomic molecules can be
accurately determined.
Why?
Explain why the
bond
enthalpy
of

a molecule is usually defined
in terms
of
a gas-phase reaction.
Why
are bond-breaking
processes always endothermic and bond-forming processes
always exothermic?
Problems
8.72
8.73
8.74
8.75
From
the following data, calculate the average bond enthalpy for
the
N-H
bond:
NH3(g)
-_.
NH2(g) + H(g)
NHz(g)
• NH(g) + H(g)
NH(g)
• N(g) + H(g)
For
the reaction
i1HO
= 435 kJ/mol
i1W

= 381
kllmol
i1W
= 360
kllmol
i1HO
=
-107.2
kllmol
Calculate the average bond enthalpy
in
0 3.
The
bond enthalpy
of
Fz(g) is 156.9 kllmol. Calculate
i1H?
for F(g).
For
the reaction
(a)
Predict the enthalpy
of
reaction from the average
bond
enthalpies in Table 8.6. (b) Calculate the enthalpy
of
reaction
from the standard enthalpies
of

formation (see Appendix 2)
of
the
reactant and product molecules, and
compare
the result with your
answer for
part
(a).
8.76 Classify the following substances as ionic compounds
or
covalent
compounds containing discrete molecules:
CH
4
,
KF, CO, SiCI
4
,
BaCl
z
·
8.77
8.78
8.79
8.80
8.81
8.82
8.83
8.84

Which
of
the following are ionic compounds and which are
covalent compounds:
RbCI,
PF
s
,
BrF
3'
KOz, CI
4
?
Match each
of
the following energy changes with
one
of
the
processes given: ionization energy, electron affinity,
bond
enthalpy, and standard enthalpy
of
formation.
(a) F(g)
+ e •
F-(g)
(c) Na(g) • Na+(g) + e-
(b) F
2

(g)
• 2F(g) (d) Na(s) +
~F2(g)
• NaF(s)
The
formulas for the fluorides
of
the third-period elements
are NaF, MgF2' AIF
3
,
SiF
4
,
PF
s
,
SF
6
,
and CIF
3
. Classify these
compounds as covalent
or
ionic.
Use
ionization energy (see Figure 7.8) and electron affinity (see
Figure 7.10) values to calculate the energy change (in
kJ/mol) for

the following reaction
s:
(a) Li(g) + leg)
+.
Li
+(g) +
r-(g)
(b) Na(g) + F(g) • Na+(g) +
F-(g)
(c) K(g) + CI(g) • K +(g) +
cqg)
Describe so
me
characteristics
of
an ionic
compound
such as
KF
that would distinguish it from a covalent
compound
such as
benzene (C
6
H
6
) .
Write Lewis structures for
BrF
3'

CIF
s
,
and
IF
7
. Identify those in
which the octet rule is not obeyed.
Write three reasonable resonance structures for the azide
ion
N3
in
which the atoms are arranged as NNN.
Show
formal charges.
The
amide
group plays an important role in determining the
structure
of
protei ns:
• •
'0'
oo
II
-N-C-
I
H
Draw
another resonance structure for this group. Show formal

c;harges.
8.85 Give an example
of
an ion
or
molecule containing Al that
(a) obeys the octet rule, (b) has an expanded octet, and (c)
ha
s
an incomplete octet.
8.86
Draw
four reasonable resonance structures for the
P0
3
F
z
-
ion.
8.87
8.88
8.89
8.90
8.91
8.92
8.93
8.94
8.95
.96
.97

The
central P atom is bonded to the three 0 atoms and to the F
atom.
Show formal charges.
Attempts
to
prepare the compounds
CF
2
,
Li0
2
,
CsCI
2
,
and PIs as
stable species under atmospheric conditions have failed.
Suggest
possible reasons for the failure.
Draw
reasonable resonance structures for the following ions:
(a)
HS0
4
,
(b)
PO
~-,
(c)

HS0
3
,
(d)
SO
~
-
.
Are
the following statements true
or
false? (a) Formal charges
represent an actual separation
of
charges. (b)
i1H
~
xn
can be
estimated from the
bond
enthalpies
of
reactants and products.
(c) All second-period elements obey the octet rule
in
their
compounds. (d)
The
resonance structures

of
a molecule can
be
separated from one another in the laboratory.
A rule for drawing plausible Lewis structures is that the central
atom is generally less electronegative than the surrounding atoms.
Explain why this is so.
Using the following information and the fact that the average
C-
H bond enthalpy
is
414
kllmol,
estimate the standard
enthalpy
of
formation
of
methane (CH
4
).
C(S)
+.
C(g)
2H
2
(g) • 4H(g)
i1H
~xn
=

716
kllmol
i1H':xn
= 872.8
kllmol
Based
on changes in enthalpy, which
of
the following reactions
will occur more readily?
(a) CI(g)
+
CHig)
-
_.
CH
3
CI(g) + H(g)
(b) CI(g) +
CH
4
(g) •
CH
3
(g) + HCl(g)
Which
of
the following molecules has the shortest nitrogen-to-
nitrogen bond: N
2

H
4
,
NzO,
N
z
, N
2
0
4
? Explain.
Most
organic acids can be represented as
RCOOH,
where
COOH
is
the carboxyl group and R is the rest
of
the molecule.
[For example, R is
CH
3
in acetic acid
(
CH
3
C
OOH
).

] (a) Draw a
Lewis structure for the carbox y
I group. (b) Upon ionization, the
carboxyl group is converted to the carboxylate group
(COO
-)
.
Draw
resonance structures for the carboxylate group.
Which
of
the following species are isoelectronic:
NH
r, C
6
H
6
,
CO,
CH
4
,
Nz, B3N3H
6?
The
following
is
a simplified (skeletal) structure
of
the amino

acid histidine. Draw a complete Lewis structure
of
the molecule.
N
> N
\
H-C-H
H
I
H-C-C-O
I I
H-N
0
I I
H H
The
following
is
a simplified (skeletal) structure
of
the amino
acid tryptophan. Draw a complete Lewis structure
of
the
molecule.
QUESTIONS
AND
PROBLEMS
309
H

I
H H
H
,,
/C"
I I
C C C
C-C
C-O-H
I I I I I I
/C"
/C"
/C"
H
N-H
0
I
H C N H
I
I
H
H H
8.98
The
following species have been detected in interstellar space:
(a) CH, (b)
OH, (c) C
2
,
(d) HNC, (e) HCO. Draw Lewis

structures for these species.
8.99 The amide ion (NH
2
)
is a Br\ilnsted
ba
se. Use Lewis structures to
represent the reaction between the amide ion and water.
8.100 Draw
Le
wis structures for the following organic molecules:
(a) tetrafluoroethylene (C
2
F
4
),
(b) propane (C3HS)' (c) butadiene
(CH
2
CHCHCH
2
),
(d) propyne (CH
3
CC
H), (e) benzoic acid
(C
6
H
s

COOH)
. (To draw C
6
H
s
COOH, replace an H
atom
in
ben
ze
ne
with a
COOH
group.)
8.101 The triiodide ion (1
3
)
in which the I atoms are arranged in a
straight line is stable, but the corresponding
F3 ion does not
exist. Explain.
8.102 Compare the bond enthalpy
of
F
z
with the overall energy change
for the following process:
8.103
8.104
8.105

8.106
8.107
8.108
8.109
F
2
(g) • F+(g) + F- (g)
Which is the preferred dissociation for F
2
,
energetically
speaking?
Methyl isocyanate (CH
3
NCO)
is used to make certain pesticides.
In
December
1984, water leaked into a tank containing this
substance at a chemical plant, producing a toxic cloud that killed
thousands
of
people in Bhopal, India.
Draw
Lewis structures for
CH
3NCO, showing formal charges.
The
chlorine nitrate (CIONO
z

)
molecule is believed to
be
in
volved in the destruction
of
ozone in the Antarctic stratosphere.
Draw a plausible Lewis structure for this molecule.
Several resonance structures for the molecule CO
2
are shown
here. Explain why so
me
of
them
are likely to be
of
little
importance in describing the bonding in this molecule.
• •
••
•
(a)
:P=C=o.:
(c)
:O=C-o.:
• •
••
••
(b)

:O-C-O:
• •
(d)
:O-C-O:
• • • •
For each
of
the following organic molecules draw a Lewis
structure in which the carbon atoms are
bonded
to each other by
single
bond
s:
(a) C
Z
H
6
,
(b) C
4
H
JQ,
(c) C
S
H
12
. For parts (b) and (c),
show only structures in which each C atom is bonded to no more
than two other C atoms.

Dr
aw Lewis structures for the following chlorofluorocarbons
(CFCs), which are partly responsible for the depletion
of
ozone in
the stratosphere: (
a)
CFCI
3
,
(b)
CF
2
Cl
z
,
(c) CHF1CI, (d)
CF
3
CHF
2
·
Dra
w
Le
wis structures for the following organic molecule
s:
C
Z
H

3
F,
C
3
H
6
,
C
4
Hs.
In
each there is one
C=C
bond, and the rest
of
the carbon atoms are
joined
by
C-C
bonds.
Calculate
i1Ho for the reaction
using (a) Equation
8
.3
and (b) Equation 5.19, given that i1H'ffor
1
2
(g) is 61.0
kl

l
mo1.
310
CHAPTER
8 Chemical
Bonding
I:
Basic Concepts
8.110 Draw Lewis
st
ructures for the following organic molecules:
(a) methanol (CH
3
0H
); (b) ethanol (CH
3
CH
z
OH); (c) tetraethyl
lead
[Pb(CH
z
CH
3
)4],
which is used in "leaded gasoline";
(d) methylamine (
CH
3
NH

2
),
which is used in tanning;
(e) mustard gas (CICH
2
CH
z
SCH
z
CH
z
CI), a poisonous gas used
in World War I;
(f) urea [(NH
z)z
CO], a fertilizer; and (g) glycine
(NHzCHzCOOH), an amino acid.
8.111 Write Lewis structures for the following four isoelectronic
species: (a)
CO, (b)
NO
+, (c)
CN
- , (d) N
2
. Show formal charges.
8.112
Oxygen forms three types
of
ionic compounds in which the

anions are oxide
(0
2
- ), peroxide
(O~
-
),
and superoxide (O
z)
.
Draw Lewis structures
of
these ions.
8.113
Comment
on the correctness
of
the statement, "A
ll
compounds
containing a noble gas atom violate the octet
rule."
8.114 Write three resonance structures for (a) the cyanate ion (NCO- )
and (b) the isocyanate ion (CNO-
).
In each case, rank the
resonance structures in order
of
increasing importance.
8.115 (a) From the following data calculate the bond enthalpy

of
the F
z
ion.
F
2
(g) • 2F(g)
t1H
~xn
= 156.9
kJ
/
mol
F- (g) • F(g) + e-
t1H
~xn
=
333kJ/moi
F z (g) • F
2
(g) + e -
t1H
~x
n
= 290 kJ/
mol
(b) Explain the difference between the bond enthalpies
ofF
z
and F

2
.
8
.11
6
The
resonance concept is sometimes described by analogy to a
mule, which is a cross between a
hor
se and a donkey. Compare
this analogy with the one used in this chapter, that is, the
description
of
a rhinoceros as a cross between a griffin and a
unicorn. Which description is more appropriate? Why?
8.117
The
N - 0 bond distance
in
nitric oxide
is
115
pm
, which is
intermediate between a triple bond (106
pm
) and a double
8.118
8.119
8.120

8.121
bond (120
pm
).
(a) Draw two resonance structures for NO, and
comment on their relative importance. (b) Is it possible to draw a
resonance structure having a triple bond between the atoms?
Although nitrogen dioxide
(NO
z
) is a stable compound, there is a
tendency for two such molecules to combine to form dinitrogen
tetroxide
(N
Z
0
4
).
Why? Draw four resonance structures
of
N
2
0
4
,
showing formal charges.
In the gas phase, aluminum chloride exists as a dimer
(a
unit
of

two) with the formula Al
z
C1
6
.
Its skeletal structure is given by
Cl Cl
CI
"'-/"'-/
AI
Al
/"'-/"'-
Cl
Cl Cl
Complete the Lewis structure and indicate the coordinate
covalent bonds in the molecule.
Draw a Lewis structure for nitrogen pentoxide
(N
2
0
S
)
in which
each N
is
bonded to three 0 atoms.
The
hydroxyl radical (OH) plays an important role in
atmospheric chemistry. It is
highly reactive and has a tendency

to combine with an H atom from other compounds, causing
them to break up. Thus
OH is sometimes called a "detergent"
radical because it helps to clean up the atmosphere. (a) Draw the
Lewis structure for the radical. (b) Refer to Table 8.6 and explain
why the radical has a
high affinity for H atoms. (c) Estimate the
entha
lp
y change for the following reaction:
OH(g) +
CHig)
+-
.
CH
3
(g) + H
2
0 (g) .
(d) The radical is generated when sunlight hits water vapor.
Calculate the maximum wavelength (in nm) required to break
an
O-H
bond in H
zO.
8.122 Vinyl chloride (C
z
H
3
CI) differs from ethylene (C

Z
H
4
)
in that one
of
the H atoms is replaced with a CI atom. Vinyl chloride is used
to prepare poly(vinyl chloride), which is an important polymer
used in pipes. (a) Draw the Lewis structure
of
vinyl chloride.
(b) The repeating unit in poly(vinyl chloride) is
-CH
2
-CHCI
Draw a portion
of
the molecule showing three such repeating
units. (c) Calculate the enthalpy change when
1.0 X 10
3
kg
of
vinyl chloride forms poly(vinyl chloride).
8.123 Experiments show that it takes 1656 kJ/mol to break all the bonds
in methane (CH
4
)
and 4006 kJ/mol to break all the bonds
in

propane (C3H
S)
'
Ba
sed on these data,
ca
lculate the average bond
enthalpy
of
the
C-C
bond.
8.124 In 1998 scientists using a special type
of
electron microscope
were able to measure the force needed to break a
single chemical
bond.
If
2.0 X 10-
9
N was needed to break a
C-Si
bond, estimate
the bond enthalpy in
kJ/mol. Assume that the bond has to be
stretched by a distance
of
2 A (2 X 10-
10

m) before it is broken.
8.125
The
American chemist Robert S. Mulliken suggested a different
definition for the electro negativity (EN)
of
an element, given by
_
IE
:
I
_
+~EA_
EN=
2
where
lEI
is the first ionization energy and
EA
is the electron
affinity
of
the element. Calculate the electronegativities
of
0,
F,
and CI using the preceding equation. Compare the
electronegativities
of
these elements on the Mulliken and Pauling

scales. (To convert
to
the Pauling scale, divide each
EN
value
by
230
kJ
/mo!.)
8.126 Among the common inhaled anesthetics are:
8.127
8.128
8.129
8.130
8.131
Halothane (CF
3
CHClBr) Isoflurane (CF
3
CHClOCHF
z
)
Enflurane (CHFCICF
z
OCHF
2
) Methoxyflurane
(CHCl
z
CF

z
OCH
3
)
Draw Lewis structures
of
these molecules. .
A student in your class claims that magnesium oxide actually
consists
of
Mg
+ and 0 - ion
s,
not
Mg
2+
and 0
2
-
ions. Suggest
some experiments one could do to show that your classmate is
wrong.
From
the lattice energy
of
KCI in Table 8.6, and the ionization
energy
of
K and electron affinity on Cl in Figures 7.8 and 7.10,
respectively, calculate the

t1H
o for the reaction
K(g) + CI(g)
+.
KCl(s
)
The species
H
~
is the simplest polyatomic ion. The geometry
of
the ion is that
of
an equilateral triangle. (a) Draw three resonance
structures to represent the ion. (b) Given the following information
2H
+ H+ •
H
~
t1W
= - 849 kJ/mol
Hz
•
2H
t1H
o = 436.4 kJ/mol
calculate t1Ho for the reaction
H+
+ H2 •
H

~
The
bond enthalpy
of
the C- N bond in the amide group
of
proteins (see Problem 8.84) can be treated as an average
of
C-N
and
C=N
bond
s.
Calculate the maximum wavelength
of
light
needed to break the bond.
In 1999 an unusual cation containing only nitrogen
(Nr)
was
prepared. Draw three resonance structures
of
the ion, showing
formal charges.
(
Hint:
The
N atoms are joined
in
a linear fashion.)

ANSWERS TO IN-CHAPTER MATERIALS
311
PRE-PROFESSIONAL PRACTICE
EXAM
PROBLEMS:
PHYSICAL
AND BIOLOGICAL SCIENCES
Nitrous oxide
(N
2
0)
is an anesthetic
commonly
used for dental procedures.
Because
of
the euphoria caused by inhaling it, N
2
0 is commonly known as
"laughing gas."
It
is licens
ed
for use as a food additive and as
an
aerosol pro-
pellant.
It
is used to displace air from potato chip
bag

s in order to extend she
lf
life and as the propellant in whipped
cream
canisters. In recent years N
2
0
has become popular as a recreational drug,
due
in
part
to its ready availabil-
ity to consumers. Although
N
2
0 is legal,
it
is regulat
ed
by the FDA; its sale
and distribution for the purpose
of
human
consumption are not permitted.
l.
Bond Bond
Enthalpy
(l,J/mol)
N-O
176

N=O
6m
N-N
193
N=N
N=N
0=0
418
94l.4
498.7
Which
of
the following Lewis structures are
possible
for N
2
0?
••
• • • •
• •
• •
:N-N-O:
'N=N=O'
• •
:N=O=N:
:N-O
-N:

• • • •
• • • •

I
II
III
IV
a) I
only
b) I
and
II
c) I, II, and
III
d) I, II,
III,
and
IV
2.
Use
formal charges to
choose
the best
of
the
re
so
nance
structures
shown.
a)
I
b) II

c)
III
d)
IV
3.
Using
the
be
st resonance structure and
the
average
bond
enthalpies
given, determine the
/::"H'ffor N
2
0.
a)
73
kJ/mol
b)
-73
kJ/mol
c) 166 kJ/
mol
d)
-166
kJ/mol
4.
Why

does the calculated /::,.Hf'value differ
from
the tabulated value
of
8l.
56
kJ/mol?
a)
The
tabulated value is wrong.
b)
None
of
the
resonance
str
uctures depicts the
bonds
realistically.
c) To a reas
onable
number
of
significant figures, the calculated and
tabulated values are the
same
.
d)
Using
bond

enthalpies gives only an
estimate
of
/::"H
~xn
'
ANSWERS
TO
IN-CHAPTER
MATERIALS
Practice Problems
8.1A
(a)
Ca
2
+,
(b)
~B~
3- , (c)
D:J

8.lE
(a) 2
-,
(b)
+,
(c) 3- .
8.2A MgCI
2
.

8.2B
NaF
<
MgO
< AIN.
8.3A
629 kJ/mo!. 8.3B
-8
41
kJ/mo!.
8.4A
(a) nonpolar, (b) polar, (c) nonpolar. 8.4B
Li
,
Na,
K,
Rb
, Cs,

'F'
• •

I

Fr, Ca, Sr,
Ba,
and Ra. 8.SA 0.12. 8.SB 0.11 D. 8.6A
:F-N-F:
• • • • • •


'0'
• •

I

8.6B
:O-CI-O:
8.7A
C
atom
= 0,
double-bonded
0 atom = 0,
• • • •
••
s
ingle-bonded
0 atoms =
-l.
8.7B S
atom
= + 1, 0
atoms
=
-1
,
• •
'0'
I I


overall
charge
=
-2
.
8.8A
-C-Q-
H
8.8B
CI-
N-Cl

• •

'0'
• •
'0'
'0'
• •
8.9A

II

:O-N-O:

I •
:O-N
=
O:
. I


:O=N-O:
•
•
•
•
• • • • • • •
• • •
o 0
-1
- 1 0 0
-?
0
+1
8.9B
[N=C-~:]-
• •
[N=C=~:]
-
· .
[B
-c
=s:]
-
8.10A
'O
- H
8.10B
:S=N-S:
8.11A

:P-Be-P:
• •
••
• • • • •
•
• • • • •
:Cl: :C]:" .
:p:


I

I
/C,r

I ;I::
8.11B
:Cl-B-Cl:
8.12A
:CI-P,
.
8.12B
:F-Sb,
.
••
••
.•
I

·

•.
j"
.Cl. .
F
:C]: . . :F: .


• •
:Cl:
••
• •
••
. . .1

:CI~I-;-CI:
8.13B :F =Kr= F:
8.14
-557
kJ.
8.13A
. . I'

••
••
• •
:Cl:

Answers
to
Checkpoints

8.1.1
b.
8.1.2
e. 8.2.1 c. 8.2.2 a. 8.4.1 b. 8.4.2 c. 8.5.1 d. 8.5.2 c. 8.6.1 e .
8.6.2
c. 8.7.1 a, b, d.
8.7.2
b.
8.8.1 e.
8.8.2
a, b, e. 8.8.3 e.
8.8.4
d. 8.9.1 a.
8.9.2
e.
Answers
to
Applying
What
You've Learned
• • • •
a)
'N'
.g.
b)
The
differen
ce
in
electronegativity is (3.5 - 3.0) =

0.5,
making
this a
polar
covalent
bond. c)
The
partial charges
on
Nand
o 0
o in
NO
are
+0
.029 and - 0.029, respectively. d)
:N=Q:
and
l ' '. '.
-
'.0
,,
-;:::-0
:
N
1+
1
H
:0:
H

I I I
H-C C C-H
I I I
:0:
H
:0:
I +1 I 1
N N +
:0
/
~
O'.
.:0
:
~
O"
• • • • • •
••
- 1 - 1
e)
Formal
charges are indicated on
each
Le
wis structure
in
part
(d).
f)
/::"H

o for the
explosive
decomposition
of
nitroglycerin
is
-l.94
X 10
3
kJ/mo!.
I
9.1
•
•
•
•
9.2
9.3
9.4
•
•
9.5
9.6
•
•
•
•
•
9.7
Molecular Geometry

The VSEPR Model
Electron-Domain Geometry
and
Molecular Geometry
Deviation from Ideal
Bond
Angles
Geometry
of
Molecules
with
More Than One
Central
Atom
Molecular Geometry
and
Polarity
Valence Bond Theory
Hybridization of Atomic
Orbitals
Hybridization
of
s
and
p Orbitals
Hybridization
of
s,
p,
and

d Orbitals
Hybridization in
Molecules Containing
Multiple Bonds
Molecular Orbital Theory
Bonding
and
Antibonding
Molecular Orbitals
(T
Molecular Orbitals
Bond
Order
TT
Molecular Orbitals
Molecular Orbital Diagrams
Bonding Theories and
Descriptions of Molecules
with Delocalized Bonding
11
lea
on
Molecular Geometry
and Bonding Theories
11
In
Mole,ular
Shape
and
Our

Sense
of
Smell
The 2004 Nobel Prize in Physiology or Medicine was awarded
to
Richard Axel and
Linda Buck for their discoveries
of
odorant receptors and the organization
of
the olfac-
tory system. Their work shed new light on how our sense of smell works. Although it is
somewhat more complicated than can be explained in detail here, the odor
of
a molecule
depends in large part on which of the olfactory receptors it stimulates. Olfactory recep-
tors are specialized proteins, located on hairlike cilia in the back
of
the nose. Stimula-
tion
of
a receptor triggers an electrical signal that travels along nerve fibers
to
the brain,
where the scent is identified. Which olfactory receptors a molecule stimulates depends
on its shape.
We
are thought to have receptors for at least seven different, fundamental scents: floral,
pepperminty, musky, pungent, camphorlike, ethereal, and putrid. Different parts of a
larger molecule may have distinctly different shapes, enabling it

to
stimulate more than
one type
of
olfactory receptor. This results in the perception of a combination
of
odors.
For example, parts
of
the benzaldehyde molecule are shaped such that they stimulate the
camphorlike, floral, and pepperminty receptors. This is a combination that we perceive
as
the smell
of
almonds. Such combinations enable us
to
recognize thousands
of
differ-
ent smells.
This discussion illustrates the importance
of
molecular geometry
to
the most enigmatic
of
our senses smell. Many biochemical processes are specific in that they depend on
the shapes
of
the molecules involved. Chemical bonding theories help

us
to
predict and/
or explain these shapes.
o~
H
Benzaldehyde
In This Chapter, You Will Learn how to determine the three-dimensional shape
of
a molecule and how the inter-
actions
of
atomic orbitals give rise
to
chemical bonds.
Before you begin, you should review
•
•
•
Shapes
of
atomic orbitals
[
~~
Section
6.7]
Electron configurations
of
atoms
[

~~
Section
6.8]
Drawing Lewis structures
[
~~
Section
8.5]
Volatile aromatic compounds give coffee its characteristic rich
aroma. Molecules responsible for the odor
of
a substance have ·
distinctive shapes, which enables them to stimulate our sense
of
smell.

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